Boolean Algebra Flashcards

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1
Q

What are the AND (∧) identities?

A

A ∧ 0 ⇒ 0
A ∧ 1 ⇒ A

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2
Q

What are the OR (∨) identities?

A

A ∨ 0 ⇒ A
A ∨ 1 ⇒ 1

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3
Q

What are the negation identities?

A

A ∧ ¬A ⇒ 0
A ∨ ¬A ⇒ 1

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4
Q

what is the absorption law?

A

A ∧ (A ∨ B) ⇒ A
A ∨ (A ∧ B) ⇒ A

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5
Q

What is the distribution law?

A

A ∧ (B ∨ C) ⇒ (A ∧ B) ∨ (A ∧ C)
(like multiplying out brackets)

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6
Q

what are De Morgan’s laws?

A

¬(A ∨ B) ⇒ ¬A ∧ ¬B
¬(A ∧ B) ⇒ ¬A ∨ ¬B

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7
Q

What is a half adder?

A

A circuit which adds two single bit binary numbers together

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8
Q

what are the input and outputs of a half adder?

A

Inputs:
- two single bit binary numbers (eg, A and B)
Outputs:
- S - the sum of the two numbers
- C - the carry bit

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9
Q

what is the logic diagram for a half adder composed of?

A

S = A XOR B
C = A AND B

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10
Q

How can you tell if a circuit is a half adder?

A

it has 2 inputs

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11
Q

what is a full adder?

A

a circuit used to add larger binary numbers together.

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12
Q

What are the inputs and outputs of a full adder?

A

Inputs:
- two single bit binary numbers (eg, A and B)
- the carry input bit (Cin)
Outputs:
- the Sum - S
- the carry output bit (Cout)

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13
Q

how can you tell a circuit is a full adder?

A

it has 3 inputs

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14
Q

what is a full adder made up of?

A

two half adders connected by an OR gate

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15
Q

what is a D-Type flip flop?

A

A circuit which stores one bit of data

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16
Q

what are the inputs and outputs of a D-type flip flop?

A
  • data input
  • clock input
  • output of Q and NOT Q
  • Q is set to the data input on the rising edge of the clock and stays that value for one clock cycle
17
Q

How do you use a Karnaugh map to simplify Boolean expressions?

A
  • split expressions by the ORs and add 1s where each section is true
  • draw boxes that are powers of 2 (eg, 1, 2, 4 etc 1s in them)
  • look at each box and see which variables stay the same
  • put each expression from each box together with ORs
18
Q

what is important to remember for Boolean simplification?

A

AND takes precedence over OR
so:
A ∧ B ∨ C ⇒ (A ∧ B) ∨ C